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Multiple Choice Questions

A circle is inscribed in an equilateral triangle of side of length l The area of any square inscribe...........

Quantitative Aptitude (CDS) Coordinate Geometry

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A circle is inscribed in an equilateral triangle of side of length l. The area of any square inscribed in the circle is

$\small \frac{l^{2}}{2}$

$\small \small \frac{\sqrt{3} l^{2}}{4}$

$\small \frac{l^{2}}{4}$

$\small \small \frac{l^{2}}{6}$

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Explanation:

Since ABC is an equilateral triangle BL = l /2

In $\small \Delta$ ABL , AB² = AL² + BL²

l²= AL² + (l/2)²

$\small AL=\frac{\sqrt{3}}{2}l$

$\small AO + OL =\frac{\sqrt{3}}{2}l......................(1)$

Also AO:OL=2:1 (property of equilateral triangle, O is the center of the circle)

AO = 2 x OL

Put above value in =n(1)

$\small 2 OL \: +\: OL=\frac{\sqrt{3}}{2}l$

$\small 3\: OL=\frac{\sqrt{3}}{2}l$

$\small OL=\frac{l}{2\sqrt{3}}$

$\small OL=OR= \frac{1}{2}\: SQ ( OL,OR\: are\: radius\: and\: SQ\: is\: diameter)$

$\small \frac{l}{2\sqrt{3}}=\frac{1}{2}QS$

So other diagonal PR is also of same length:

$\small PR=\frac{l}{\sqrt{3}}$

$\small Arear PQRS = \frac{l}{2}\times QS \times PR$

$\small =\frac{1}{2}\times \frac{l}{\sqrt{3}} \times \frac{l}{\sqrt{3}}$

$\small =\frac{l^{2}}{6}$

By: Munesh Kumari

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